Saudi Booklet

Geometry Difficulty 5.6 AIME, harder Prove it Saudi Arabia

Given two circles $(O), (O')$ with different radii and intersecting at $A, B$ . Circle of center $A$ , radius $AB$ intersects $(O), (O')$ again at $C, D$ respectively. Let $EF$ be the common tangent closer to $B$ of the two circles with $E \in (O), F \in (O')$ . Rays $AE, AF$ intersect $BC, BD$ at $M, N$ respectively. Prove that the internal bisector of $\angle CBD$ passes through the circumcenter of triangle $AMN$ .

Solution

Since $AB = AC$ , in circle $(O)$ , we have $\angle ABC = \angle AEB$ so $ABM, AEB$ are two similar triangles. This implies that
$\frac{AM}{AB} = \frac{AB}{AE} \implies AB^2 = AM \cdot AE.$
Similarly, one could get $AB^2 = AN \cdot AF$ so $AM \cdot AE = AN \cdot AF$ , thus $EMNF$ is cycle. Hence, $\angle AMN = \angle AFE$ . On the other hand
$\angle AFE = \angle AFB + \angle BFE = \angle ABD + \angle BAF = \angle BNF$
so $\angle AMN = \angle BNF = \angle AND$ , this means that $BD$ in tangent to $(AMN)$ . Similarly, $BC$ is also tangent to $(AMN)$ , so the center of $(AMN)$ is equidistant from the two segments $BC$ and $BD$ . In other words, the angle bisector of $\angle CBD$ passes through the center of $(AMN)$ . $\square$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.